This question concerns bit strings of length six. These bit strings can be divided up into four types depending on their initial and terminal bit. Thus the types are: 00, 01, 10, 11.

How many bit strings of length six must you select before you are sure to have at least 6 that are of the same type? (Assume that when you select bit strings you always select different ones from ones you have already selected.)

You must select at least 21 strings of length 6.

Step-by-step explanation:

There are 2⁴ = 16 strings of each type.

Since there are 4 types of strings, then in order to have 6 strings of the same type you need to have at least 21 strings selected. The worst case will that there are 5 strings for each of the other 3 types.

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: